Precalculus Final Exam Review 1

A vertical asymptote is found by setting the simplified _________= to zero and solving for x  (a)  

• Numerator
• Denominator
• Right side 
• Left side

Tags

CCSS.HSF-IF.C.7D

What are the asymptotes?

• x = -3        y = 1  
• x = 3         y = 1
• x = -3        y = -1
• x = 3         y = -1

Tags

CCSS.HSF-IF.C.7D

• y = -7

• x = 7

• y = 0

• x = -7

Tags

CCSS.HSF-IF.C.7D

Label the graph

b

c

d

e

a

Vertical Asymptote

y-intercept

origin

Horizontal Asymptote

x-axis

Tags

CCSS.HSF-IF.C.7D

What is the x-intercept?

• x= -5

• x= 5

• x= 6

• x= -6

Tags

CCSS.HSF-IF.C.7D

Where are the holes in the function?

• x = 4

• x = -4

• x = 2

• x = -2

Tags

CCSS.HSF-IF.C.7D

Which type of conic section is this equation?

• circle

• ellipse

• hyperbola

• parabola

Tags

CCSS.HSG.GPE.A.1

Which type of conic section is this equation?

• circle

• ellipse

• hyperbola

• parabola

Tags

CCSS.HSG.GPE.A.1

Which type of conic section is this equation?

• circle

• ellipse

• hyperbola

• parabola

Tags

CCSS.HSG.GPE.A.1

Write the equation of the circle with a radius of 3 and a center of (4, 2). 

• ( x - 4)² + (y - 2)² = 3
• ( x - 2)² + (y - 2)² = 3²
• ( x - 4)² + (y - 2)² = 3²
• ( x + 4)² + (y + 2)² = 3²

Tags

CCSS.HSG.GPE.A.1